Problem 39
Question
Construct the molecular orbital diagram for CF. Would you expect the bond length of \(\mathrm{CF}^{+}\) to be longer or shorter than that of CF?
Step-by-Step Solution
Verified Answer
Yes, the bond length of \(\mathrm{CF}^{+}\) is expected to be longer than that of CF.
1Step 1: Identify the Atomic Orbitals (AOs)
Starting with the carbon atom which is in the second period, Carbon (C) has 1s, 2s, and 2p atomic orbitals. Fluorine (F), also in the second period, has the same 1s, 2s, and 2p atomic orbitals.
2Step 2: Overlap the AOs to form MOs
Overlap these atomic orbitals to form molecular orbitals. Atomic s orbitals overlap to form sigma molecular orbitals, while p orbitals can form both sigma and pi orbitals. Remember that each pairwise overlap generates one bonding and one antibonding orbital. This would give us \(\sigma_{2s}\), \(\sigma_{2s}^*\), \(\sigma_{2p}\), \(\pi_{2p}\), \(\pi_{2p}^*\), and \(\sigma_{2p}^*\) molecular orbitals.
3Step 3: Fill the MOs with Electrons
Using the Aufbau principle (fill from lowest orbital up) and the Pauli exclusion principle (maximum of two electrons per orbital with opposite spins), populate the molecular orbitals. CF has 6 valence electrons from C and 7 from F, for a total of 13 valence electrons (\(C:[He]2s^{2}2p^{2}\), F:\([Ne]2s^{2}2p^{5}\)). In CF+\, we have one less electron, so that's 12 valence electrons. The order of filling would be \(\sigma_{2s}\), \(\sigma_{2s}^*\), \(\sigma_{2p}\), \(\pi_{2p}\), \(\pi_{2p}^*\), and \(\sigma_{2p}^*\).
4Step 4: Interpret the Impact on Bond Length
A decrease in the number of bonding electrons or an increase in the number of antibonding electrons would increase bond length. Here, we notice that losing an electron from CF to form CF+ involves removing a bonding electron, implying a lesser bond attraction and therefore a longer bond length in CF+ compared to CF.
Key Concepts
Understanding Atomic OrbitalsBond Length in Molecular StructuresElectron Configuration and Molecular Orbitals
Understanding Atomic Orbitals
Atomic orbitals (AOs) are the regions around an atom's nucleus where electrons are most likely to be found. Each atom has its own set of atomic orbitals, which is determined by its electron configuration.
For example, carbon and fluorine, both in the second period of the periodic table, have the same types of atomic orbitals: 1s, 2s, and 2p. These orbitals define how electrons are distributed around the atom.
For example, carbon and fluorine, both in the second period of the periodic table, have the same types of atomic orbitals: 1s, 2s, and 2p. These orbitals define how electrons are distributed around the atom.
- **1s Orbital**: Closest to the nucleus and fully occupied by two electrons.
- **2s Orbital**: Also spherical but larger, can hold up to two electrons.
- **2p Orbital**: Dumbbell-shaped and holds up to six electrons across three sub-orbitals.
Bond Length in Molecular Structures
Bond length is the distance between the nuclei of two bonded atoms. It gives us essential insights into the strength and stability of the bond.
Factors affecting bond length include:
Factors affecting bond length include:
- **Number of Bonding vs. Antibonding Electrons**: More bonding electrons generally lead to shorter bonds, since positive synergism within bonds brings nuclei closer.
- **Bond Order**: Higher bond order typically means a shorter bond length due to more sharing of electron density between nuclei.
Electron Configuration and Molecular Orbitals
The electron configuration of atoms tells us how electrons are arranged in atomic orbitals. According to molecular orbital theory, when these atoms form molecules, their atomic orbitals combine to create molecular orbitals (MOs).
For CF, we first look at the electron configuration of carbon \[ \text{(C): [He]2s^{2}2p^{2} } \] and fluorine \[ \text{(F): [Ne]2s^{2}2p^{5} } \]. These atoms come together, with carbon's and fluorine's atomic orbitals overlapping, to form several MOs: \- **Sigma (\(\sigma\)) and Sigma Star (\(\sigma^*\))**: Primarily formed by overlapping s orbitals and some p orbitals. - **Pi (\(\pi\)) and Pi Star (\(\pi^*\))**: Created from the side-to-side overlap of p orbitals.
Electrons fill from the lowest MO to the highest, following the Aufbau principle. For CF+, one fewer electron results in a specific electron configuration within these MOs, affecting the overlap strength and thus the bond length. Understanding which orbitals are occupied allows us to predict molecular properties like bond order and stability.
For CF, we first look at the electron configuration of carbon \[ \text{(C): [He]2s^{2}2p^{2} } \] and fluorine \[ \text{(F): [Ne]2s^{2}2p^{5} } \]. These atoms come together, with carbon's and fluorine's atomic orbitals overlapping, to form several MOs: \- **Sigma (\(\sigma\)) and Sigma Star (\(\sigma^*\))**: Primarily formed by overlapping s orbitals and some p orbitals. - **Pi (\(\pi\)) and Pi Star (\(\pi^*\))**: Created from the side-to-side overlap of p orbitals.
Electrons fill from the lowest MO to the highest, following the Aufbau principle. For CF+, one fewer electron results in a specific electron configuration within these MOs, affecting the overlap strength and thus the bond length. Understanding which orbitals are occupied allows us to predict molecular properties like bond order and stability.
Other exercises in this chapter
Problem 37
Consider the molecules \(\mathrm{NO}^{+}\) and \(\mathrm{N}_{2}^{+}\) and use molecular orbital theory to answer the following: (a) Write the molecular orbital
View solution Problem 38
Consider the molecules \(\mathrm{CO}^{+}\) and \(\mathrm{CN}^{-}\) and use molecular orbital theory to answer the following: (a) Write the molecular orbital con
View solution Problem 40
Construct the molecular orbital diagram for CaF. Would you expect the bond length of \(\mathrm{CaF}^{+}\) to be longer or shorter than that of CaF?
View solution Problem 41
Explain why the concept of delocalized molecular orbitals is essential to an understanding of bonding in the benzene molecule, \(\mathrm{C}_{6} \mathrm{H}_{6}\)
View solution